A small generalization. Any doubly transitive action of a group $G$ on a set $X$ has the property that $\frac{1}{|G|} \sum_{g \in G} \text{Fix}(g)^2 = 2$. This is because, by double transitivity, the diagonal action of $G$ on $X^2$ has precisely two orbits, the orbit where the first factor equals the second and the orbit where it doesn't, so the result follows by Burnside's lemma. Similarly, any $k$-transitive action of a group $G$ on a set $X$ with $k < |X|$ has the property that $\frac{1}{|G|} \sum_{g \in G} \text{Fix}(g)^k = B_k$, since the orbits of the diagonal action of $G$ on $X^k$ can naturally be put into bijection with partitions of a $k$-element set. For $k \ge |X|$ the sum evaluates to the number of partitions of a $k$-element set into at most $|X|$ parts (of course, the action can only be $k$-transitive for $k = |X|$ if $G$ is in fact the full group of permutations on $X$!).
The representation-theoretic upshot of all this is that the permutation representation corresponding to a doubly transitive group action always breaks down into the direct sum of one copy of the trivial representation and an irreducible representation. This is an easy way to write down nice irreducible representations of groups like $PSL_2(\mathbb{F}_q)$, which has a triply transitive action on $\mathbb{P}^1(\mathbb{F}_q)$. \mathbb{P}^1(\mathbb{F}_q)$ (edit: when $q$ is even!)
